Chapter 4, Problem 4.112QP

(a)

Interpretation Introduction

Interpretation:

The net ionic equation for the reaction of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ with $\text{NaOH}$ should be written.

Concept introduction:

Acid -Base reaction:

The reaction between acid and base to produce a salt is called acid-base reaction.

Net ionic equation:

The only species which are contributes in the reaction are written in form of ions in reaction is called net ionic equation.

Molarity:

The concentration of the solutions is given by the term molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

The mole of compound is given by the ratio between taken mass of the compound and molar mass of the compound.

$\text{Mole}\text{=}\frac{\text{Mass of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

(b)

Interpretation Introduction

Interpretation:

The percent purity of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ in given sample should be calculated.

Concept introduction:

Acid -Base reaction:

The reaction between acid and base to produce a salt is called acid-base reaction.

Net ionic equation:

The only species which are contributes in the reaction are written in form of ions in reaction is called net ionic equation.

Molarity:

The concentration of the solutions is given by the term molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

The mole of compound is given by the ratio between taken mass of the compound and molar mass of the compound.

$\text{Mole}\text{=}\frac{\text{Mass of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$